A string under a tension of 129.6 N produces | vedclass

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(A) The frequency of a vibrating string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass dens

Vedclass · Accepted Answer

$100$

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(A) The frequency of a vibrating string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Let $f_0$ be the frequency of the tuning fork.Case $1$: Tension $T_1 = 129.6 \ N$. The string produces $10 \ beats/s$,so $f_1 = f_0 \pm 10$.Case $2$: Tension $T_2 = 160 \ N$. The string is in unison,so $f_2 = f_0$.Since $T_2 > T_1$,$f_2 > f_1$,which implies $f_1 = f_0 - 10$.Thus,$f_0 - f_1 = 10 \Rightarrow f_0 - \frac{1}{2l} \sqrt{\frac{129.6}{\mu}} = 10$.Also,$f_0 = \frac{1}{2l} \sqrt{\frac{160}{\mu}}$.Dividing the two equations: $\frac{f_0}{f_0 - 10} = \sqrt{\frac{160}{129.6}} = \sqrt{\frac{1600}{1296}} = \frac{40}{36} = \frac{10}{9}$.$9f_0 = 10f_0 - 100$.$f_0 = 100 \ Hz$.

$A$ string under a tension of $129.6 \ N$ produces $10 \ beats/s$ when it is vibrated along with a tuning fork. When the tension in the string is increased to $160 \ N$,it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork in $Hz$.

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